Rationalise the denominators of the following:
(i) \(\frac{1 }{ \sqrt{7 }}\)
(ii) \(\frac{1 }{ \sqrt{7 }-\sqrt6}\)
(iii) \(\frac{1 }{ \sqrt{5}+\sqrt2}\)
(iv) \(\frac{1 }{ \sqrt{7}-2}\)
(i) \( \frac{1}{\sqrt{7}} \)
To rationalize the denominator, multiply both the numerator and denominator by \( \sqrt{7} \):
\[ \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7} \] So, the rationalized form of \( \frac{1}{\sqrt{7}} \) is \( \frac{\sqrt{7}}{7} \).
(ii) \( \frac{1}{\sqrt{7} - \sqrt{6}} \)
To rationalize the denominator, multiply both the numerator and denominator by \( \sqrt{7} + \sqrt{6} \) (the conjugate of \( \sqrt{7} - \sqrt{6} \)):
\[ \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} \] Simplifying: \[ = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6} \] So, the rationalized form of \( \frac{1}{\sqrt{7} - \sqrt{6}} \) is \( \sqrt{7} + \sqrt{6} \).
(iii) \( \frac{1}{\sqrt{5} + \sqrt{2}} \)
To rationalize the denominator, multiply both the numerator and denominator by \( \sqrt{5} - \sqrt{2} \) (the conjugate of \( \sqrt{5} + \sqrt{2} \)):
\[ \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} \] Simplifying: \[ = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3} \] So, the rationalized form of \( \frac{1}{\sqrt{5} + \sqrt{2}} \) is \( \frac{\sqrt{5} - \sqrt{2}}{3} \).
(iv) \( \frac{1}{\sqrt{7} - \sqrt{2}} \)
To rationalize the denominator, multiply both the numerator and denominator by \( \sqrt{7} + \sqrt{2} \) (the conjugate of \( \sqrt{7} - \sqrt{2} \)):
\[ \frac{1}{\sqrt{7} - \sqrt{2}} \times \frac{\sqrt{7} + \sqrt{2}}{\sqrt{7} + \sqrt{2}} = \frac{\sqrt{7} + \sqrt{2}}{(\sqrt{7})^2 - (\sqrt{2})^2} \] Simplifying: \[ = \frac{\sqrt{7} + \sqrt{2}}{7 - 2} = \frac{\sqrt{7} + \sqrt{2}}{5} \] So, the rationalized form of \( \frac{1}{\sqrt{7} - \sqrt{2}} \) is \( \frac{\sqrt{7} + \sqrt{2}}{5} \).
For real number a, b (a > b > 0), let
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \leq a^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi\)
and
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \geq b^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi\)
Then the value of (a – b)2 is equal to _____.